Q.Assertion: Ethanol is a weaker acid than phenol.
Reason: Sodium ethoxide may be prepared by the reaction of ethanol with aqueous NaOH.
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Start your 14-day free trial to unlock the full solution →The assertion -- ethanol is a weaker acid than phenol -- is correct, because the phenoxide ion is resonance-stabilised while the ethoxide ion is not. But the reason -- that sodium ethoxide can be prepared by reacting ethanol with aqueous NaOH -- is factually wrong: aqueous NaOH cannot effectively deprotonate ethanol, because water is itself a slightly stronger acid than ethanol. The correct option is (iii).
Why This Question Tests Conceptual Depth
This assertion-reason question checks two separate things: whether you know ethanol is a weaker acid than phenol, and whether you know how sodium ethoxide is actually prepared.
1. The assertion: ethanol vs. phenol acidity
Acidity depends on the stability of the conjugate base after losing H+.
- Phenol () loses H+ to form the phenoxide ion (), whose negative charge is delocalised into the aromatic ring by resonance. This stabilisation makes phenol give up its proton comparatively easily.
- Ethanol () loses H+ to form the ethoxide ion (), where the charge stays localised on oxygen -- no resonance is possible.
Relative acidity: Phenol (pKa ~10) is stronger than Ethanol (pKa ~16). The assertion is correct.
2. The reason: can sodium ethoxide be made from ethanol + aqueous NaOH?
This part is wrong. Consider the equilibrium:
CH3CH2OH + NaOH <=> CH3CH2ONa + H2O
For this to proceed usefully to the right, hydroxide would have to be a better base than ethoxide -- i.e. water would have to be a weaker acid than ethanol. The opposite is true: water's pKa (~15.7) is slightly lower than ethanol's (~16), meaning water is the marginally stronger acid and hydroxide the weaker base. So the equilibrium lies almost entirely on the reactant side -- aqueous NaOH does not give a useful yield of sodium ethoxide. …
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